Two plates (area = S) charged to +q1 and +q2(q2

Question

Two plates (area = S) charged to +q1 and +q2(q2 < q1) brought closer to form a capacitor of capacitance C. The potential difference across the plates is: (A)  (q1-q2/2C)  (B)  (q1-q2/C)  (C)  (q1-q2/4C)  (D)  (2(q1-q2)/C)

Tardigrade Admin · Accepted Answer

Capacitance of the parallel plate capacitor is

C = \frac{ε _{0} A}{d}

where

A

is area of plates and

d

is distance between them.
Potential difference across the plates

V = E d

Here,

E = \frac{( q _{1} - q _{2} )}{2 ε _{0} A}

∴ V = \frac{( q _{1} - q _{2} ) d}{2 ε _{0} A}

= \frac{q _{1} - q _{2}}{2 C} .

Q. Two plates (area = $S$ ) charged to $+ q_{1}$ and $+ q_{2} (q_{2} < q_{1})$ brought closer to form a capacitor of capacitance $C$ . The potential difference across the plates is:

$\frac{q _{1} - q _{2}}{2 C}$

$\frac{q _{1} - q _{2}}{C}$

$\frac{q _{1} - q _{2}}{4 C}$

$\frac{2 ( q _{1} - q _{2} )}{C}$

Q. Two plates (area = S) charged to +q1​ and +q2​(q2​<q1​) brought closer to form a capacitor of capacitance C. The potential difference across the plates is:

2Cq1​−q2​​

Cq1​−q2​​

4Cq1​−q2​​

C2(q1​−q2​)​

Capacitance of the parallel plate capacitor is C=dε0​A​ where A is area of plates and d is distance between them. Potential difference across the plates V=Ed Here, E=2ε0​A(q1​−q2​)​ ∴V=2ε0​A(q1​−q2​)d​ =2Cq1​−q2​​.

Q. Two plates (area = $S$ ) charged to $+ q_{1}$ and $+ q_{2} (q_{2} < q_{1})$ brought closer to form a capacitor of capacitance $C$ . The potential difference across the plates is:

$\frac{q _{1} - q _{2}}{2 C}$

$\frac{q _{1} - q _{2}}{C}$

$\frac{q _{1} - q _{2}}{4 C}$

$\frac{2 ( q _{1} - q _{2} )}{C}$